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AS Physics Unit 7 Forces in Equilibrium Mr D Powell 7 Forces in Equilibrium The addition of vectors by calculation or scale drawing. Calculations will be limited to two perpendicular vectors. The resolution of vectors into two components at right angles to each other; examples should include the components of forces along and perpendicular to an inclined plane. Conditions for equilibrium for two or three coplanar forces acting at a point; problems may be solved either by using resolved forces or by using a closed triangle. Moment of a force about a point defined as force × perpendicular distance from the point to the line of action of the force; torque. Couple of a pair of equal and opposite forces defined as force × perpendicular distance between the lines of action of the forces. The principle of moments and its applications in simple balanced situations. Centre of mass; calculations of the position of the centre of mass of a regular lamina are not expected. Mr Powell 2009 Index Chapter Map Mr Powell 2009 Index 7.1 Vectors and Scalars 1. What is a vector quantity? 2. How do we represent vectors? 3. How do we add and resolve vectors? Representing a vector Addition of vectors using a scale diagram Addition of two perpendicular vectors using a calculator Resolving a vector into two perpendicular components Spec 3.2.1 Mr Powell 2009 Index A Definition.... Scalars Vectors mass, temperature, time, length, speed, energy displacement, force, velocity, acceleration, momentum TASK: Make a note of each quantity, then look at the units and give a reason why they have been sorted as such.. Mr Powell 2009 Index Displacement and velocity A runner completes one lap of an athletics track. What distance has she run? 400 m What is her final displacement? If she ends up exactly where she started, her displacement from her starting position is zero. What is her average velocity for the lap, and how does it compare to her average speed? Mr Powell 2009 Index What did you decide? Despite the trip moving at various speeds, because it ended up at the starting point, the average velocity was zero. This will always be true when the final displacement is zero. Mr Powell 2009 Index Vector equations An equation is a statement of complete equality. The left hand side must match the right hand side in both quantity and units. In a vector equation, the vectors on both sides of the equation must have equal magnitudes and directions. Take Newton’s second law, for example: force = mass × acceleration Force and acceleration are both vectors, so their directions will be equal. Mass is a scalar: it scales the right-hand side of the equation so that both quantities are equal. Force is measured in newtons (N), mass in kilograms (kg), and acceleration in ms-2. The units on both sides must be equal, so 1 N = 1 kgms-2. Mr Powell 2009 Index Displacement vectors Harry and Sally are exploring the desert. They need to reach an oasis, but choose to take different routes. Harry travels due north, then due east. N Sally simply travels in a straight line to the oasis. When Harry met Sally at the oasis, they had travelled different distances. However, because they both reached the same destination from the same starting point, their overall displacements were the same. Mr Powell 2009 Index Vector notation A scalar quantity is often represented by a lower case letter, e.g. speed, v. A vector quantity can also be represented by a lower case letter, but it is written or printed in one of the following formats to differentiate it from the scalar equivalent: The value of a vector can be written in magnitude and direction form: e.g. v = (v, θ) y 6 a Or as a pair of values called components: e.g. a = (8, 6) or 8 8 x 6 Mr Powell 2009 Index Vector addition Displacement vectors can always be added ‘nose to tail’ to find a total or resultant vector. y b This can be done approximately by scale drawing: a a+b 10 x 7 It can also be done by calculation, breaking each vector down into perpendicular components first and then adding these together to find the components of the resultant: c+d = 2 3 + -2 2 = 0 5 Mr Powell 2009 Index Calculating a resultant When adding two perpendicular vectors, it is often necessary to calculate the exact magnitude and direction of the resultant vector. This requires the use of Pythagoras’ theorem, and trigonometry. For example, what is the resultant vector of a vertical displacement of 3 km and a horizontal displacement of 4 km? magnitude: R2 = 32 + 42 R 4 km R = √ 32 + 42 = √ 25 θ direction: tan θ = 4/3 θ = tan-1(4/3) = 53° = 5 km 3 km Mr Powell 2009 Index Vector components Just as it is possible to add two vectors together to get a resultant vector, it is very often useful to break a ‘diagonal’ vector into its perpendicular components. This makes it easier to describe the motion of an object, and to do any relevant calculations. vertical component horizontal component Mr Powell 2009 Index Calculating components A vector can be separated into perpendicular components given only its magnitude and its angle from one of the component axes. This requires the use of trigonometry. For example, what are the horizontal and vertical components of a vector with a magnitude of 6 ms-1 and a direction of 60° from the horizontal? cos60° = x / 6 x = 6 × cos60 sin60° = y / 6 y = 6 × sin60 6 ms-1 y = 3 ms-1 = 5.2 ms-1 60° x Mr Powell 2009 Index Pythagoras & Adding Vectors TASK: Now try out the same techniques for two more triangles; 1) opp = 2m, adj = 8m Answers 1. 2. = 14 , R =8.25m = 20.6 , R =4.27m 2) opp = 1.5m, adj = 4m Mr Powell 2009 Index Applications... TASK: Now try out the same techniques for two more triangles; 1) What is R & if opp = 54N, adj = 22N Answers 1. 2. = 67.83 , R =58.3m =39.4 , R =65.6ms-1 2) What is adj & if opp = 54ms-1, R = 85ms-1 Mr Powell 2009 Index Splitting a vector into components... TASK: Put on your maths brain and try and explain using trig why Vx is related to cos and Vy is related to sin Mr Powell 2009 Index Resolving in a context.... Answers TASK: 1) A bird is flying at a speed of 15ms-1 at an angle of 36 to the horizontal. What is its vertical speed. 1. Vy =8.8ms-1 2. Vy =70.7N Vx = 70.7N 2) A javelin is thrown at a resultant force of 100N at a 45 angle. What are the horizontal and vertical forces on it. Mr Powell 2009 Index Questions to try on your own.... Answers ..... Mr Powell 2009 Index More Trig Practice 1. How many radians in a circle? 2. How many radians in quarter of a circle? 3. Work out in degrees for a right-angled triangle if hyp = 5, opp = 4, adj = 3? 4. Work out opp if = 450 and the adj = 7cm? 5. Work out tan if opp = 53.0m, adj = 42.0m? 6. Work out opp if = /2 and the adj = 3cm? 7. Work out in radians for a right-angled triangle if hyp = 5, opp = 4, adj = 3? 8. Rearrange sin=opp/hyp to make the subject. 9. Work out sin (/2) =, sin (2) = , cos () = , cos(cos-1(/4)) = , tan(/4) = 10.Use formulae & numbers from your calculator to prove that; opp tan adj Mr Powell 2009 Index Vector Problems 1 Question 1 Both sides contribute as vectors so double it to If using only pythag do simply 2 x 15N Sin20 = 10.26 = 10.3N label upper part of triangle as a. Hence a2 = (152+152-2x152Cos140)0.5 = 28.19N, then half for triangle. To 14.1N. Then pythag (152-14.12)0.5 = 5.12N (x2) = 10.23N a b c Mr Powell 2009 Index Vector Problems 1 Question 2 Mr Powell 2009 Index Vector Problems 1 Question 3 3) A mass of 20.0kg is hung from the midpoint P of a wire. Calculate the tension in each suspending wire in Newton’s. Assume g = 10ms-2. (Hint resolve…..) Use the idea of point of equilibrium. The forces must balance vertical and horizontal. Weight balances the tension so for each wire Use the 2T Cos70 = 200 T Cos70 = 100 T = 100 / Cos70 = 292.38N Mr Powell 2009 Index Vector Problems 1 Question 5/6 5) Magnitude = sqrt(35002+ 2792) = 3511.1m = 3510m Direction = tan-1 (279/3500) = 4.557° = 4.6° south from vertical 3500km 6) Vertical motion: 21 sin (43°) = 14.32 ms-1 Horizontal motion: 21 cos (43°) =15.358 ms-1 S = 3 x 15.358 ms-1 = 46.1 ms-1 0.279km Mr Powell 2009 Index Vector Problems 2 Question 2 Use the parallelogram method to resolve the forces acting on this object placed. (Hint employ both cosine & sine rule). (4 marks) Mr Powell 2009 Index Vector Problems 2 Question 2 Use the parallelogram method to find the resultant acting on this object and angle. (Hint employ both cosine & sine rule). (4 marks) Mr Powell 2009 Index Vector Problems 2 Question 3 3) Two forces of magnitude 10.0N and F Newton’s produce a resultant of magnitude 30.0N in the direction OA. Find the direction and magnitude of F. (2 marks) (Hint use Pythagoras) Mr Powell 2009 Index Vector Problems 2 Question 4 The graph shows a part completed vector diagram. You task is to find out the vector R by mathematical analysis. The vectors A & B are shown in both coordinate and bearing formats. Show working for both a mathematical method (2 marks) and drawn out scaled method. (2 marks) Hint: this is not as hard as it might initially look! Mr Powell 2009 Index Vector Problems 2 Question 4 - stage 1 Finding the components of vectors for vector addition involves forming a right triangle from each vector and using the standard triangle trigonometry. The vector sum can be found by combining these components and converting to polar form. Mr Powell 2009 Index Vector Problems 2 Question 4 – stage 2 After finding the components for the vectors A and B, and combining them to find the components of the resultant vector R, the result can be put in polar form by Mr Powell 2009 Index Vector Problems 2 - Question 5 Mr Powell 2009 Index 5a N 5b N N 6) Forces of 60.0N and F Newton’s act at point O. Find the magnitude and direction of F if the resultant force is of magnitude 30.0N along OX (2 marks) Vector Problems 2 - Question 7 Forces of 60.0N and F Newton’s act at point O. a c SinA SinC Find the magnitude and direction of F if the resultant force is of magnitude 30.0N along OX (2 marks) c2 = a2 + b2 - 2abCosC Mr Powell 2009 Index Using the Sine or Cosine Rule 30N In this problem we can solve it with other methods of trigonometry. Make a triangle of vectors where we are trying to find length c and angle A; b = 30N 120 A a c SinA SinC a = 60N c=F Sine Rule Cosine Rule Use reciprocal version of above c2 = a2 + b2 - 2abCosC c2 = 60N2 + 30N2 – 2x60Nx30N xCos120 c2 = 4500N2 +1800N2 c2 = 6300N c = 79.4N c = F = 79.4N SinC / c = SinA / a (60N x Sin 120) / 79.4N) = SinA Sin-1 (60N x Sin 120) / 79.4N) =A A = 40.876 = 41 Mr Powell 2009 Index 7) Simultaneous Equations (A*+) If this problem is to be solved mathematically we must establish two formulae. One in the vertical and one in the horizontal; 30N The key is knowing that the resultant force is 30N at 0º. Hence; Resolving Horizontally; 30N = FCos + 60NCos(360-120) 30N = FCos + 60N x -0.5 30N = FCos - 30N 60N = FCos Eq 1 Resolving Vertically; 0 = FSin + 60NSin(360-120) 0 = FSin + 60NSin(240) 0 = FSin + 60N x 3/2 0 = FSin -51.96N 51.96N = FSin Eq 2 So by dividing Eq 2 by Eq 1 FSin / FCos = 51.95N / 60N tan = 0.866 = 40.89 F = 60N / cos = 79.4N Or F = 51.96N / sin = 78.96N Mr Powell 2009 Index Wires Example - Correct So by turning it on its side we can use the same formulae (which is confusing but correct) Left side Vertically 1.20N = T1cos30+ T2cos60 1.20N = 0.866T1+ 0.5T2 Horizontally T1sin30 = T2sin60 0.5T1= 0.866T2 T1= 1.732T2 Eq 2 Eq 1 right side Sub 2 into 1; 1.20N = 0.866T1+ 0.5T2 1.20N = 0.866 x 1.732T2 +0.5T2 1.20N = 1.99T2 T2 = 0.60N & T1= 1.732T2 T1= 1.732 x 0.6N T1= 1.04N Mr Powell 2009 Index WRONG! Vertically 1.20N = T1sin30+ T2sin60 1.20N = 0.5T1+ 0.866T2 Horizontally T1cos30 = T2cos60 0.866T1= 0.5T2 T1= 0.577T2 Eq 2 Eq 1 Sub 2 into 1; 1.20N = 0.866T1+ 0.5T2 1.20N = 0.866 x 0.577T2 +0.5T2 1.20N = 1T2 T2 = 1.2N & T1= 0.577T2 T1= 0.577x1.2N T1= 0.69N Mr Powell 2009 Index Alternative Can do by using (90- ) as angle; If we take the outside angle instead of the inner angle we can do this and use the triangle; (90-) Vertically 1.20N = T1sin60+ T2sin30 1.20N = 0.866T1+ 0.5T2 Horizontally T1cos60 = T2cos30 0.5T1= 0.866T2 T1= 1.732T2 Eq 2 Eq 1 Sub 2 into 1; 1.20N = 0.866T1+ 0.5T2 1.20N = 0.866 x 1.732T2 +0.5T2 1.20N = 1.99T2 T2 = 0.60N & T1= 1.732T2 T1= 1.732 x 0.6N T1= 1.04N Mr Powell 2009 Index 7.2 Balanced Forces 1. Why do we have to consider the direction in which a force acts? 2. When do two (or more) forces have no overall effect on a point object? 3. What is the parallelogram of forces? Equilibrium of a point object Testing three forces in equilibrium Practical - Coplanar forces Mr Powell 2009 Index Why are the objects not moving... Mr Powell 2009 Index Equilibrium Point of an Object Mr Powell 2009 Index Example 1 Mr Powell 2009 Index Example 1 Mr Powell 2009 Index Example 2 Mr Powell 2009 Index Practical... Assemble the equipment shown below. Use some weights of different types and measure the angles shown. The basic system will adjust in angle to whatever masses you add. The tension in each string must equal the force exerted by W1 or W2. This enables the problem to be solved by resolving Try it practically so get a range of angles & weights then prove the equilibrium. Mr Powell 2009 Index Example 3 Mr Powell 2009 Index Example 1... 1. If you think that F1 & F2 must resolve and their components in H & V directions must balance with the weights to make the point P be in equilibrium. 2. Define the problem as shown below with two new angles and triangles. Make sure you use the correct angles! Using the following figures the problem is very simple... W1 = 5N W2 = 3N W3 = 6N 1 = 150 2 = 124 F1sin F1 F2 F2sin F2cos F1cos Mr Powell 2009 Index Example 1... V 0 Using the following figures the problem is very simple... F1 sin F2 sin W3 F3 4.33 1.67 6 W1 = 5N W2 = 3N W3 = 6N 1 = 150 2 = 124 =60 = 34 H 0 F1 cos F2 cos 2.5 2.5 F1sin F1 F2 F1cos F2sin F2cos Mr Powell 2009 Index Plenary Question A 2kg chicken rests on point C on clothes line ACB as shown. What is the minimum breaking strength of the line to ensure the bird does not break the line? H 0 V 0 F1 cos30 F2 cos 45 F1 sin 30 F2 sin 45 20N 3 2 F1 F2 2 2 2 F1 F2 3 F1 2 F2 20N 2 2 F1 F2 2 40N F1sin30 F2 F1 30 F1cos 30 F2sin45 45 F2cos45 Mr Powell 2009 Index Plenary Question A 2kg chicken rests as point C on clothes line ACB as shown. What is the minimum breaking strength of the line to ensure the bird does not break the line? F1 F2 2 40N F1 F2 F1 F2 2 40N F1 F2 F2 2 3 2 F2 2 40N 3 F2 3F2 3 40N 2 34.64 F2 12.68N 2.732 2 3 F1 F2 2 3 2 F1 12.68N 10.35N 3 So we have now worked out both the tensions in the line that are there from the weight of the chicken. This means that the line must be of a strength equal or better than F2 = 12.68N F2= 12.68N F1=10.35N F2sin45 F1sin30 30 F1cos 30 45 F2cos45 Mr Powell 2009 Index Practical Results… Weight +/0.01N 1.96 2.45 2.94 3.43 3.92 4.41 4.91 ave Sin (ave) In terms of forces Hyp = F opp = mg Hence... opp/hyp = adj mg/F=sin(theta) Angle +/-1 deg 24 24 24 24 24 26 27 24.71 0.42 Weight +/-0.01N 6.00 y = 0.4905x R² = 1 5.00 Weight Added (N) Force on Newton mass +/-1g meter +/-0.2N 4.0 200 5.0 250 6.0 300 7.0 350 8.0 400 9.0 450 10.0 500 4.00 3.00 2.00 1.00 Fsin(theta) = mg F=x m = sin(theta) y = mg 0.00 0.0 2.0 4.0 6.0 8.0 10.0 Force on meter (N) Mr Powell 2009 Index 12.0 Resolution of Forces – ICT Link Self Check http://www.walter-fendt.de/ph14e/forceresol.htm Mr Powell 2009 Index 7.3 Principle of Moments 1. Under what conditions does a force produce a turning effect? 2. How can the turning effect of a given force be increased? 3. What is required to balance a force that produces a turning effect? 4. Why is the centre of mass an important idea? Turning effects, The principle of moments, Centre of mass, Centre of mass tests, Calculating the weight of a metre rule Mr Powell 2009 Index Centres of mass and gravity The centre of gravity of an object is a point where the entire weight of the object seems to act. The centre of mass of an object is a point where the entire mass of the object seems to be concentrated. In a uniform gravitational field the centre of mass is in the same place as the centre of gravity. An alternative definition is that the centre of mass or centre of gravity of an object is the point through which a single force has no turning effect on the body. Mr Powell 2009 Index Equilibrium A body persists in equilibrium if no net force or moment acts on it. Forces and moments are balanced. Newton’s first law states that a body persists in its state of rest or of uniform motion unless acted upon by an external unbalanced force. Bodies in equilibrium are therefore bodies that are at rest or moving at constant velocity (uniform motion). F1 F2 F2 F1 equilibrium Mr Powell 2009 Index The principle of moments The principle of moments states that (for a body in equilibrium): total clockwise moments total anticlockwise moments = This principle can be used in calculations: 5m What is d? d 4 × 5 = 6d 4N 6N 20 = 6d d = 20 / 6 d = 3.3 m Mr Powell 2009 Index Human forearm The principle of moments can be used to find out the force, F, that the biceps need to apply to the forearm in order to carry a certain weight. When the weight is held static, the system is in equilibrium. weight of arm = 20 N 60 N Taking moments about the elbow joint: 4 cm F schematic diagram 4F = (16 × 20) + (35 × 60) 16 cm 20 N 4F = 2420 60 N 35 cm F = 605 N Mr Powell 2009 Index Finding the weight of a metre rule The uniform metre rule shown is in equilibrium, with its centre of gravity marked by the arrow ‘weight’. Find the weight of the metre rule. 0.2 m 0.3 m 0.5 m W 3N total anticlockwise moments = total clockwise moments 3 × 0.2 = weight × 0.3 weight = 0.6 / 0.3 weight = 2 N Mr Powell 2009 Index Work out the mass then Density of a ruler... Mr Powell 2009 Index Example Results + errors! For a wooden ruler a density is quoted at 750kgm-3 d1 = (0.2 +/- 0.001)m +/- 0.5% d2 = (0.2 +/- 0.001)m +/- 0.5% m = (96+/1)g = (0.096 +/- 0.001)kg +/- 0.1% Thickness = (0.0061 +/- 0.00001)m +/- 0.16% Width = (0.028 +/- 0.00001)m +/- 0.04% Value for our ruler density... Length = (1 +/- 0.001)m +/- 0.1% m1xd1 = m2xd2 Error in mass = 1.1% (562.1 +/- 8) kgm-3 So this must be a different wood! Density = m/v Error in density = 1.1%+ 0.16% + 0.04% + 0.1% = 1.4% Mr Powell 2009 Index 7.4 More on Moments 1. What can we say about the support force on a pivoted body? 2. When a body in equilibrium is supported at two places, how much force is exerted on each support? 3. What is meant by a couple? Support forces - Single support problems - Two support problems Couples ISA preparation: practical task – practice in recording measurements and plotting a graph written task Section A and B – opportunities to discuss reliability and to analyse data and errors Mr Powell 2009 Index Practical Investigating the bridge crane Aims 1. To use simple measurements of distance and to apply the principle of moments to a model bridge crane. 2. To relate practical measurements to straight line graph theory. 3. To consider measurement errors and how to reduce them. You require the following equipment: 1. 2. 3. 4. 5. 6. 7. 8. two metre rulers two spring balances calibrated in newton a set square slotted masses of weight that can be measured using either balance thread scissors two stands and clamps graph paper Safety Ensure the practical arrangement is stable and will not topple over. Mr Powell 2009 Index Setup Advice.... Ensure that the spring balances are vertical and that the ruler is horizontal each time before you make their measurements. D You may need to use a set square and a second metre ruler to do this. Use straight line graph theory and the general equation; y = mx + c Theory.... Taking moments about the point where spring balance S2 supports the bar gives: S1D = Wd2 + 0.5W0D S1 = Wd2/D + 0.5W0 Y = S1 , x = d2 , m = W/D c = 0.5Wo Mr Powell 2009 Index Setup Advice.... S1D = Wd2 + 0.5W0D S1 = Wd2/D + 0.5W0 Ensure that the spring balances are vertical and that the ruler is Y = S1 , x = d2 , m = W/D c = 0.5Wo horizontal each time before they make their measurements. You may need to use a set square and a second metre ruler to do this. D Use straight line graph theory and the general equation; y = mx + c It is recommended that you carry out calculations in SI units in order to avoid confusion. Results W = (6N +/- 1.8) N Wo = (1.4 +/- 0.42)N The weight W should be equal to the gradient of the line D. The weight W0 should be equal to 2 the y-intercept. Mr Powell 2009 Index Answers to Questions 1 (a) Use the second metre ruler to measure the vertical height (above the bench) of the horizontal ruler at each end. Use the set square against the bench (assumed horizontal) to check the second metre ruler is vertical. Adjust the clamps holding the horizontal ruler if necessary and recheck it is horizontal. Use the set square to check the spring balances are vertical. (b) The spread of the points about the line of best fit gives an indication of reliability. If the points are spread too much about the line of best fit, the results unreliable and the measurements should be repeated. 2) (a) Ensuring the ruler is horizontal and the spring balances are vertical each time the support forces S1 and S2 are to be measured. Precision of Newton meter. (b) Repeat each measurement at least twice for each position of the weight and calculate the mean value of each support force. Use more precise instrument with better scale. 3) (a) The spring balance does not read zero when it is unloaded. (b) The gradient would be the same. The y-intercept would be different. Mr Powell 2009 Index Couples... If two forces act about a hinge in opposite directions, there is an obvious turning effect called a couple. The resulting linear force from a couple is zero. The couple is given by the simple formula: G = 2 Fs This strange looking symbol, G, is “gamma”, a Greek capital letter ‘G’. Couples are measured in Newton metres (Nm) The turning effect is often called the torque. It is a common measurement made on motors and engines, alongside the power. Racing engines may be quite powerful but not have a large amount of torque. This is why it would not make sense for a racing car to be hitched to a caravan, any more so than trying to win a Formula 1 race in a 4 x 4. Mr Powell 2009 Index Couples and torques A couple is a pair of forces acting on a body that are of equal magnitude and opposite direction, acting parallel to one another, but not along the same line. Forces acting in this way produce a turning force or moment. The torque of a couple is the rotation force or moment produced. The forces on this beam are a couple, producing a moment or torque, which will cause the beam to rotate. F d F Mr Powell 2009 Index The torque of a couple There is a formula specifically for finding the torque of a couple. A point P is chosen arbitrarily. Take moments about P. F P x total moment = Fx + F(d – x) d–x F d = Fx + Fd – Fx = Fd torque of a couple = force × perpendicular distance between lines of action of the forces Mr Powell 2009 Index 7.5 Stability 1. What is the difference between stable and unstable equilibrium? 2. When is a tilted object going to topple over? 3. Why is a vehicle more stable, the lower its centre of mass? Mr Powell 2009 Index Mr Powell 2009 Index Forces Triangles... Resolving Forces Parallel to Slope F = W x sin hyp x sin = opp hyp x cos = adj Resolving Perpendicular to Slope Sx + Sy = W x cos Mr Powell 2009 Index 7.6 Equilibrium Rules 1. What condition must apply to the forces on an object in equilibrium? 2. What condition must apply to the turning effects of the forces? 3. How can we apply these conditions to predict the forces acting on a body in equilibrium? Free body force diagrams The triangle of forces The conditions for equilibrium of a body Mr Powell 2009 Index Visuals… Mr Powell 2009 Index Quick Test what does each one represent... 1. A point on an object where the entire weight of the object seems to act. In a uniform gravitational field, it is in the same place as the centre of mass. 2. A point on an object where the entire mass of the object seems to be concentrated. In a uniform gravitational field, it is in the same place as the centre of gravity. 3. A pair of forces acting on a body that are of equal magnitude and opposite direction, acting parallel to one another, but not along the same line. 4. The state a body is in if no net force or moment acts on the body. 5. The point about which a lever turns. Also called the pivot. 6. The turning effect of a force or forces. Can also be called the torque (symbol τ). It is calculated by multiplying the force by the perpendicular distance between the pivot and the line of action of the force. The units of moment are newton metres (Nm). 7. The point about which a lever turns. Also called the fulcrum. 8. The principle stating that the sum of the clockwise moments are equal to the sum of the anticlockwise moments acting on a body. 9. The turning effect of a force or forces, given the symbol τ. Can also be called the moment. It is calculated by multiplying the force by the perpendicular distance between the pivot and the line of action of the force. The units of torque are newton metres (Nm). 10. The rotation force or moment produced by a couple. It can be calculated by multiplying the force by the perpendicular distance between the lines of action of the forces. Its units are newton metres (Nm). 11. The force created by the gravitational attraction on a mass. Its units are newtons (N). Mr Powell 2009 Index Quick Test Answers... 1. centre of gravity – A point on an object where the entire weight of the object seems to act. In a uniform gravitational field, it is in the same place as the centre of mass. 2. centre of mass – A point on an object where the entire mass of the object seems to be concentrated. In a uniform gravitational field, it is in the same place as the centre of gravity. 3. couple – A pair of forces acting on a body that are of equal magnitude and opposite direction, acting parallel to one another, but not along the same line. 4. equilibrium – The state a body is in if no net force or moment acts on the body. 5. fulcrum – The point about which a lever turns. Also called the pivot. 6. moment – The turning effect of a force or forces. Can also be called the torque (symbol τ). It is calculated by multiplying the force by the perpendicular distance between the pivot and the line of action of the force. The units of moment are newton metres (Nm). 7. pivot – The point about which a lever turns. Also called the fulcrum. 8. principle of moments – The principle stating that the sum of the clockwise moments are equal to the sum of the anticlockwise moments acting on a body. 9. torque – The turning effect of a force or forces, given the symbol τ. Can also be called the moment. It is calculated by multiplying the force by the perpendicular distance between the pivot and the line of action of the force. The units of torque are newton metres (Nm). 10. torque of a couple – The rotation force or moment produced by a couple. It can be calculated by multiplying the force by the perpendicular distance between the lines of action of the forces. Its units are newton metres (Nm). 11. weight – The force created by the gravitational attraction on a mass. Its units are newtons (N). Mr Powell 2009 Index Exam Question ...Jan 07 Spec A Q5 5 (a) Define the moment of a force. (b) The diagram shows the force, F, acting on a bicycle pedal. (i) The moment of the force about O is 46Nm in the position shown. Calculate the value of the force, F. (2 marks) (ii) Force, F, is constant in magnitude and direction while the pedal is moving downwards. State and explain how the moment of F changes as the pedal moves through 80°, from the position shown. (2 marks) Mr Powell 2009 Index Exam Question ...Jan 07 Spec A Q5 5 (a) Define the moment of a force. (b) The diagram shows the force, F, acting on a bicycle pedal. (i) The moment of the force about O is 46Nm in the position shown. Calculate the value of the force, F. (2 marks) (ii) Force, F, is constant in magnitude and direction while the pedal is moving downwards. State and explain how the moment of F changes as the pedal moves through 80°, from the position shown. (2 marks) Mr Powell 2009 Index Exam Question ...Jan 07 Spec A Q5 5b)…. (i) The moment of the force about O is 46Nm in the position shown. Calculate the value of the force, F. (2 marks) (ii) Force, F, is constant in magnitude and direction while the pedal is moving downwards. State and explain how the moment of F changes as the pedal moves through 80°, from the position shown. (2 marks) 46Nm 240N Mr Powell 2009 Index